How do you use atoi to assign individual elements of a char array?

Question

Hi,

So as we all probably know, the atoi converts a char to a number. But, what do you do if you only want one of the array elements instead of the whole array?

Please look at the following:

for (h = 0; h < 5; h++)
{
    num[h] = atoi(temp[h]);
}

Assume that num is an array of type int and that temp is and array of type char. This gives me one of those annoying conversion problems:

Invalid conversion from 'char' to 'const char *'

Any suggestions on how to convert a single element of a char array to an int using atoi?

Answer 1

If you only want to convert a single character you don't need to use atoi():

if (temp[h] >= '0' && temp[h] <= '9')
{
    num[h] = temp[h] - '0';
}
else
{
    // handle error:  character was not a digit
}

In C, the value of each digit is one greater than the value of the previous digit, so this is guaranteed to work.

The reason that atoi() does not work is because it takes a const char* as its argument, not a char. That pointer has to point to a null terminated string.

Answer 2

Besides just using its integral value as shown by James, you could put it in a seperate buffer:

char buf[2] = { temp[h], '\0' };
num[h] = atoi(buf);